MATH 110 Module 3 Exam _Statistics (Portage learning)

Stati sti cs - Portage Online Summer 2018
Module 3 Exam
Exam Page 1
Suppose A and B are two events with probabilities:
P(Ac )=.40,P(B)=.45,P(A∪B)=.60.
Find the following:
a) P(A∩B).
p(anb) = p(a) + p(b) - p(aub)
p(a) = 1 - p(a^c)
p(a^c) = 0.40
1 - 0.40 = 0.60 p(a) = 0.60 0.60 + 0.45 - 0.60 = 0.45 P(AnB) = 0.45
b) P(A).
p(a) = 1 - p(a^c)
p(a^c) = 0.40
1 - 0.40 = 0.60 P(A) = 0.60 c) P(Bc).
p(b) = 1 - p(b^c)
rearranged to find p(b^c) = 1 - p(b) 1 - 0.45 = 0.55 This study source was downloaded by 100000843865143 from CourseHero.com on 01-20-2023 19:21:38 GMT -06:00
https://www.coursehero.com/file/32320826/Module-3-Examdocx/ Stati sti cs - Portage Online Summer 2018
P(B^c) = 0.55 Answer Key
Suppose A and B are two events with probabilities:
P(Ac )=.40,P(B)=.45,P(A∪B)=.60.
Find the following:
a) P(A∩B).
For P(A∩B). Use P(A ∪B)=P(A)+P(B)-P(A∩B) and rearrange to
P(A∩B)=P(A)+P(B)-P(A ∪B). But for this equation, we need P(A) which we can find by using P(A)=1-
P(A^c ). So, P(A)=1-.40= .60.
P(A∩B)=.60+.45-.60=.45.
b) P(A).
P(A) was found above as .60.
c) P(Bc).
For P(Bc ). Use P(B)=1-P(Bc ) which may be rearranged to (Bc )=1-P(B).
P(Bc )=1-.45=.55.
Exam Page 2
Suppose you are going to make a password that consists of 4 characters chosen from {2,7,8,c,f,k,t,z}. How many different passwords can you make if you cannot use any character more than once in each password?This study source was downloaded by 100000843865143 from CourseHero.com on 01-20-2023 19:21:38 GMT -06:00
https://www.coursehero.com/file/32320826/Module-3-Examdocx/